Inequalities in one Variable MCQ Quiz in বাংলা - Objective Question with Answer for Inequalities in one Variable - বিনামূল্যে ডাউনলোড করুন [PDF]

Concept:

Rules for Operations on Inequalities:

• Adding the same number to each side of an inequality does not change the direction of the inequality symbol.
• Subtracting the same number from each side of an inequality does not change the direction of the inequality symbol.
• Multiplying each side of an inequality by a positive number does not change the direction of the inequality symbol.
• Multiplying each side of an inequality by a negative number reverses the direction of the inequality symbol.
• Dividing each side of an inequality by a positive number does not change the direction of the inequality symbol.
• Dividing each side of an inequality by a negative number reverses the direction of the inequality symbol.

Calculation:

Given:

Let us rewrite the given inequality, we get

⇒ 15x

⇒ 15x

⇒ 15x - 16x

⇒ -x

⇒ x > 4

Therefore, we can say that the solution set for the given inequality is .

Calculation:

We have, 17 - (2x + 4) ≤ 9x - 4(2x - 3)

⇒ 17 - 2x - 4 ≤ 9x - 8x + 12

⇒ 17 - 4 - 12 ≤ 9x - 8x + 2x

⇒ 1 ≤ 3x

⇒ x ≥ 

∴ The solution set is x ∈   .

Calculation

Given;

Inequality: 37 - (3x + 5) ≥ 9x - 8(x - 3)

⇒ 37 - 3x - 5 ≥ 9x - 8x + 24

⇒ 32 - 3x ≥ x + 24

⇒ 32 - 24 ≥ x + 3x

⇒ 8 ≥ 4x

⇒ 4x ≤ 8

⇒ x ≤ 2

∴ The solution set is (-∞, 2].

Hence option 3 is correct.

Calculation

Given:

⇒ 

⇒ 

⇒  x \geq \frac{11}{-3}\)

⇒   x \geq -\frac{11}{3}\)

⇒ 

∴ x lies in the interval

Hence option 1 is correct

Calculation:

We have, 17 - (2x + 4) ≤ 9x - 4(2x - 3)

⇒ 17 - 2x - 4 ≤ 9x - 8x + 12

⇒ 17 - 4 - 12 ≤ 9x - 8x + 2x

⇒ 1 ≤ 3x

⇒ x ≥ 

∴ The solution set is x ∈   .

Concept:

If |x| ≤ a then - a ≤ x ≤ a

Calculations:

Given , |3x - 5| ≤ 2 

⇒ - 2 ≤ 3x - 5 ≤ 2

⇒ - 2 + 5 ≤ 3x  ≤ 2 + 5 

⇒ 3 ≤ 3x  ≤ 7

⇒ 

Hence, if |3x - 5| ≤ 2 then then 

Calculation:

We have, 17 - (2x + 4) ≤ 9x - 4(2x - 3)

⇒ 17 - 2x - 4 ≤ 9x - 8x + 12

⇒ 17 - 4 - 12 ≤ 9x - 8x + 2x

⇒ 1 ≤ 3x

⇒ x ≥ 

∴ The solution set is x ∈   .

Explanation:

Given system of linear inequality is 

3x - 7 \\\ 11 - 5x \le 1 \end{matrix} \right.\)

When 5 + x > 3x - 7 ⇒ 2x

When 11 - 5x ≤ 1 ⇒ 5x ≥ 10 ⇒ x ≥ 2

On the number line, the solution is represented as below.

Concept:

If |x| ≤ a then - a ≤ x ≤ a

Calculations:

Given , |3x - 5| ≤ 2 

⇒ - 2 ≤ 3x - 5 ≤ 2

⇒ - 2 + 5 ≤ 3x  ≤ 2 + 5 

⇒ 3 ≤ 3x  ≤ 7

⇒ 

Hence, if |3x - 5| ≤ 2 then then 

Concept:

Modulus function: It gives the absolute value of a variable (x). It is denoted by |x|

• |x| ≥ 0 for x ∈ R

For any real number a:

• |x| ≤ a  - a ≤ x ≤ a
• |x| ≥ a  - x ≤ -a or x ≥ a

Calculation:

We have, ||x - 2| - | = 3

⇒ |x - 2| -  = ±3

⇒ |x - 2| = 3 +  or -3 + 

Now, for real values of x,

3 +  ≥ 0 or -3 +  ≥ 0

⇒ k ≥ - 6 or k ≥ 6

∴ k ∈  or k ∈ 

Since  is included in the interval .

∴ k ∈ .

The correct answer is Option 2.